# Thread: Find the minimum value

1. ## Find the minimum value

Find the minimum value of (5 cosh x + 3 sinh x).

How can I solve this?

2. Originally Posted by geton
Find the minimum value of (5 cosh x + 3 sinh x).

How can I solve this?
Differentiate, put to zero, test nature .... the usual deal.

Note that to solve when the derivative is equal to zero, you should:

1. Replace cosh and sinh with their exponential definitions.
2. Multiply through by 2 e^{-x}.
3. Solve the resulting quadratic equation in e^x .....

3. $\displaystyle \cosh x = \frac{e^x + e^{-x}}{2}$
$\displaystyle \sinh x = \frac{e^x - e^{-x}}{2}$

so, $\displaystyle f(x) = 5\cosh x + 3 \sinh x$

$\displaystyle f'(x) = 5\sinh x + 3\cosh x = \frac{8e^x - 2e^{-x}}{2} = 4e^x - e^{-x} = \frac{4e^{2x} - 1}{e^x}$

we want to find $\displaystyle x$ such that $\displaystyle f'(x) = 0$
since $\displaystyle e^x$ is never zero, we will solve $\displaystyle 4e^{2x} - 1 = 0$.. solving, we shall obtain

$\displaystyle x = -\ln 2$... substitute that $\displaystyle x$ to find the minimum value..

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to check, $\displaystyle f''(x) = f(x) = 5\cosh x + 3 \sinh x = 4e^x + e^{-x} > 0$ for any $\displaystyle x$.
thus, $\displaystyle x = -\ln 2$ gives the minimum value for $\displaystyle f(x)$

4. Yes I got x = - ln2.

I've no idea how can I get this.

5. Originally Posted by geton
Yes I got x = - ln2.

I've no idea how can I get this.
Yes, the correct answer is 4.

$\displaystyle f(-\ln 2)=4$

-ln(2) is not the minimum value of 5cosh(x)+3sinh(x), it's just x such that the function is at its minimum.

Got the difference ?

6. Originally Posted by Moo
Yes, the correct answer is 4.

$\displaystyle f(-\ln 2)=4$

-ln(2) is not the minimum value of 5cosh(x)+3sinh(x), it's just x such that the function is at its minimum.

Got the difference ?

Yup

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# max value of sechx

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